$(\lambda,D)$-model homogenity

Question

Here on the page $41$ in the definition $6.1(3)$ I do not follow where $D$ from $6.1(3)$ appears in the definition of $(\lambda,D)$-model homogenity in $6.1(2)$. It appears in the first $2$ paragraphs in the definition $6.1$ but not in the $\lambda-$sequence-homogenity from $6.1(2)$ so that $D=D(M)$ in $6.1(3)$ makes sense via its appearance in item $6.1(2)$.

Answer 1

Turning my comments into an answer. First we fix some set $D$, which is a set of types (without parameters). For a model $M$ the notation $D(M)$ means the set of all types (without parameters) realised in $M$. So this has nothing to do with the $D$ we fixed before.

Definition 6.1 consists of three parts, which are as follows.

1. Here we define the class $K_D$. This is the class of all models $M$ such that $D(M) = D$. That is, the class of all models that only realise types in $D$.
2. This is the usual definition of $\lambda$-homogeneity. Roughly saying that any partial elementary map with small domain can be extended to any small set ("small" meaning $< \lambda$). The definition in the paper only talks about extending by one element at a time, but a simple induction construction shows we can extend the domain of that partial elementary map to any set of cardinality $< \lambda$. This definition has nothing to do with $D$ or $D(M)$ or any of that.
3. The definition of $(\lambda, D)$-homogeneous. The point here is that these are the homogeneous models in the class $K_D$ (from part 1). So a model $M$ is $(\lambda, D)$-homogeneous if it is $\lambda$-homogeneous (in the sense of part 2) and $D(M) = D$.

In particular if we would start with some model $M$ and pick $D = D(M)$ then such a model is $(\lambda, D)$-homogeneous precisely when it is $\lambda$-homogeneous. Because the second requirement (that $D(M) = D$) is then trivially satisfied.

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The point of this construction is to only consider models that realise certain types. Put differently, we will only be interested in models that omit certain types (namely the types not in $D$). This is interesting when for example studying vector spaces. Let $T$ be the first-order theory of real vector spaces. That is, we have two sorts $F$ and $V$, and $T$ specifies that $F$ is a real closed field and $V$ is a vector space over $F$. Of course there will be models where $F$ is not (isomorphic to) the standard reals $\mathbb{R}$. We do not want to consider those models. So we would be interested in studying the class of models where $F \cong \mathbb{R}$.

We can do that as follows. Add a constant for every element of $\mathbb{R}$ to our theory and add all the formulas that those elements satisfy in $\mathbb{R}$ (also called the elementary diagram of $\mathbb{R}$). Consider the following partial type (where $x$ is of sort $F$): $$ \Sigma(x) = \{x \neq r : r \in \mathbb{R} \}. $$ Now let $D$ be the set of all types not containing $\Sigma$. Then $K_D$ will be the class of all real vector spaces. Because by construction we made sure that every model of contains at least a copy of $\mathbb{R}$ (by adding the elementary diagram of $\mathbb{R}$). Any model $M$ where $F$ is bigger than $\mathbb{R}$ will satisfy a type that contains $\Sigma$. So for such a model we have $D(M) \neq D$, and hence $M \not \in K_D$.